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Karo-lina-s [1.5K]
3 years ago
6

Two technicians are discussing replacing heater hoses. Technician A says that the replacement hoses should be the same length as

the original hoses. Technician B says that the replacement hoses should be cut as short as possible to allow more coolant to flow through the heater core faster. Which technician is correct?
A. Technician A only
B. Technician B only
C. Both Technician A and B
D. Neither Technician A nor B
Physics
1 answer:
Eva8 [605]3 years ago
4 0

Answer:

A

Explanation:

In the given case, the replacing of heater hose is being discussed.

Technically technician A is correct and technician B is wrong. This is because the length of the heater hose cannot be shortened. If done so, there would be chances of busting of the hose because of the excess of heat developed in the hose. ( The heat transfer rate would decrease).

A. Technician A only

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Answer:

D

Explanation:

The decrease in potential energy is equal to the increase in kinetic energy.

mgh

250 x 9.8 x 30

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4 0
3 years ago
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The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

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The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0
3 years ago
How much force is needed to stop a body of mass 10kg​
almond37 [142]

Answer:

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Explanation:

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3 0
2 years ago
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